74
457 次の等式をすべて満たす2次関数f(x) を求めよ。
S²,₁ f(x)dx=0, ²f(x)dx=10, ₁₁xf(x) dx =
fix = ax²+bx+c & dico (ato)
こ
S²₁ fix dx = 2 Só (ax + c) dx = 2 [3x³ + cx] !
2x (=^+C)
S(0)=[
= a +2b12c
S ²₁ x fu) dx = 25, bx³² dx = 2 [x²] != = = b
b
条件から
2 (a + c) = 0, $a+20+20 = 10,
これを解と
a = 3₁ b = 2₁ C = - 1
(atoを満たす)
LT-PM, 1 f(x) = 3x²+2x-1₁₁
4
10 = 1/
3